Problem: Suppose we have a vector field $f(x, y) = (-5, y\sin(x))$ and a curve $C$ that is parameterized by $\alpha(t) = (3t, 1)$ for $-3 < t < 1$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Solution: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = (-5, y\sin(x))$ and $\alpha(t) = (3t, 1)$. $\begin{aligned} &f(\alpha(t)) = (-5, \sin(3t) ) \\ \\ &\alpha'(t) = (3, 0) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_{-3}^1 (-5, \sin(3t)) \cdot (3, 0) \, dt$ Let's solve the integral. $\begin{aligned} &\int_{-3}^1 (-5, \sin(3t)) \cdot (3, 0) \, dt \\ \\ &= \int_{-3}^1 -15 \, dt \\ \\ &= -60 \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = -60$.